merged = insert(intervals[i], merged) i = i + 1 return merged def main(): sol = Solution() i1 = Interval(2,3) i2 = Interval(4,5) i3 = Interval(2,2) i4 = Interval(15,18) intervals = [i1,i2,i3,i4] intervals = sol.
res = []whilelen(intervals)>=2:# - if overlap, update [0] and drop [1]ifintervals[1][0] <= intervals[0][1]: intervals[0][1] =max(intervals[0][1], intervals[1][1]) intervals.pop(1)else:# - put [0] into res and pop [0]res.append(intervals[0]) intervals.pop(0)# -...
class Solution: def eraseOverlapIntervals(self, intervals: list) -> int: intervalLen = len(intervals) # special considerations: if intervalLen == 0 or intervalLen == 1: return 0 # parameters scan, count = 1, 0 # sort the list to find the interval with smallest right boundary intervals ...
https://leetcode.com/problems/merge-intervals/ 排序题。 解题思路:先将区间按照每个start的值来排序,排好序以后判断一个区间的start值是否处在前一个区间中,如果在前一个区间中,那么合并;如果不在,就将新区间添加。 code :由ref改造 class Solution(object): def merge(self, intervals): """ :type inter...
end = e class Solution(object): def merge(self, intervals): """ :type intervals: List[Interval] :rtype: List[Interval] """ result = [] if not intervals: return result intervals.sort(key = lambda x: x.start) # 按照左区间排序(请看下方总结) result.append(intervals[0]) # 先将第一...
[Leetcode][python]Merge Intervals/合并区间 题目大意 给出多个数据区段,把首尾相连的数据段合并。 注意点: 所给的数据段是乱序的 解题思路 把起始位置(start)排序。遍历数据段,并与结果集中最后一个数据段比较能否合并,如果能合并就合并,否则加入结果集。
代码(Python3) class Solution: def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None: """ Do not return anything, modify nums1 in-place instead. """ # i/j 分别表示 nums1/nums2 中还未使用的最大数的下标 i, j = m - 1, n - 1 # k 表示 num...
classSolution{/*public int[][] merge(int[][] intervals) { List<int[]> merged = new ArrayList<>(); if (intervals.length == 0 || intervals == null) return merged.toArray(new int[0][]); Arrays.sort(intervals, (a, b) -> a[0] - b[0]); ...
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NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature. 题目大意: intervals列表有一些小区间,合并所有重叠的区间。 解题思路: 合并区间要考虑3种情况: [1,3] [2,6] ->[1,6] 第一个区间的end大于等于第二个区间的start且小于第...